Solve
\[\frac{x}{x + 3} \ge 0.\]Enter your answer using interval notation.
We can build a sign chart:

\[
\begin{array}{c|ccc}
& x < -3 & -3 < x < 0 & 0 < x \\ \hline
x + 3 & - & + & + \\
x & - & - & + \\
\frac{x}{x + 3} & + & - & +
\end{array}
\]Also, $\frac{x}{x + 3} = 0$ for $x = 0.$

Thus, the solution is $x \in \boxed{(-\infty,-3) \cup [0,\infty)}.$